Question

An artificial cell is made by encapsulating \(0.2 \, \text{M}\) glucose solution within a semipermeable membrane. The osmotic pressure developed when the artificial cell is placed within a \(0.05 \, \text{M}\) solution of NaCl at 300 K is ______ \( \times 10^{-1} \, \text{bar} \). (Nearest Integer)
Given: \(R = 0.083 \, \text{L bar mol}^{-1} \text{K}^{-1}\)
Assume complete dissociation of NaCl.

Answer 1

Correct Answer: 25

For NaCl dissociation:
\[ \text{NaCl} \rightarrow \text{Na}^+ + \text{Cl}^- \]
Concentration of NaCl = 0.05 M.
Effective concentration (\(C_1\)) = 0.05 M + 0.05 M = 0.1 M. 
Glucose concentration (\(C_2\)) = 0.2 M.
The osmotic pressure:
\[ \pi = (C_2 - C_1)RT \]

Substituting:
\[ \pi = (0.2 - 0.1) \times 0.083 \times 300 = 24.9 \times 10^{-1} \, \text{bar} \]
Nearest integer = 25 bar.

Answer 2

Step 1: Identify effective (osmolar) concentrations inside and outside
Inside the artificial cell: glucose solution, 0.2 M. Glucose is a non-electrolyte, so van ’t Hoff factor i = 1. Effective osmolarity inside:
C_in,osm = (0.2)(1) = 0.20 osmol L⁻¹.

Outside solution: 0.05 M NaCl, assume complete dissociation → i = 2. Effective osmolarity outside:
C_out,osm = (0.05)(2) = 0.10 osmol L⁻¹.

Step 2: Net osmotic driving concentration
Only water crosses the semipermeable membrane, so the osmotic pressure difference depends on the difference in effective concentrations:
ΔC_osm = C_in,osm − C_out,osm = 0.20 − 0.10 = 0.10 osmol L⁻¹.

Step 3: Use van ’t Hoff relation for osmotic pressure
π = R T ΔC_osm.
Given R = 0.083 L·bar·mol⁻¹·K⁻¹, T = 300 K, ΔC_osm = 0.10 mol·L⁻¹:
π = (0.083)(300)(0.10) bar = 2.49 bar.

Step 4: Express in the required format and round
2.49 bar = 24.9 × 10⁻¹ bar. Nearest integer = 25.

Final answer

25